Why Factor Classification Matters
Before working with a rational function’s denominator, identify what its factors are over the real numbers. A quadratic with real roots contributes linear factors; a quadratic with no real roots remains an irreducible quadratic factor. Repeated factors must also be counted with their full multiplicities.
These distinctions organize the denominator correctly. A factor written as a square may conceal repeated linear factors, and a quadratic that looks unfamiliar may be irreducible over the reals. Classifying the factors first prevents both errors.
A rational function \(P(x)/Q(x)\) is proper when \(P\) and \(Q\) are polynomials, \(Q\) is not the zero polynomial, and \(\deg P<\deg Q\). Here the focus is on recognizing the real factors of \(Q\) and their multiplicities, not on finding decomposition coefficients.
Linear and Irreducible Quadratic Factors
For a quadratic \(q(x)=ax^2+bx+c\), where \(a\ne 0\), the discriminant is \(\Delta=b^2-4ac\). Its sign determines how the quadratic factors over the reals:
| Discriminant | Real roots | Factor classification |
|---|---|---|
| \(\Delta>0\) | Two distinct real roots | Two distinct linear factors |
| \(\Delta=0\) | One repeated real root | A repeated linear factor |
| \(\Delta<0\) | No real roots | An irreducible quadratic factor |
The quadratic formula gives the roots when \(\Delta\ge 0\):
When the roots are \(r_1\) and \(r_2\), the quadratic factors as \(a(x-r_1)(x-r_2)\). When \(\Delta=0\), the two roots coincide, so the factorization is \(a(x-r)^2\). A negative discriminant means the quadratic has no real root and therefore no real linear factor.
Why the Discriminant Gives This Classification
Completing the square shows directly why the sign of \(\Delta\) matters. Starting with \(q(x)=ax^2+bx+c\), factor out \(a\) from the quadratic expression and complete the square:
If \(\Delta>0\), the equation \(q(x)=0\) requires the square to equal the positive number \(\Delta/(4a^2)\), so there are two real solutions. If \(\Delta=0\), the square must be zero, giving one root counted twice. If \(\Delta<0\), the expression inside the brackets is a square plus a positive number, so it cannot be zero for real \(x\). Thus the quadratic has no real roots and cannot have a real linear factor. This proves the classification.
What Multiplicity Means
A factor’s multiplicity is the number of times it occurs in the denominator’s factorization. More precisely, a polynomial factor \(F(x)\) has multiplicity \(m\) in \(Q(x)\) if \(F(x)^m\) divides \(Q(x)\), but \(F(x)^{m+1}\) does not. A nonzero constant multiplying the denominator is not counted as a factor multiplicity.
For instance, \((x-3)^4\) contains the linear factor \(x-3\) with multiplicity \(4\). Likewise, if \(x^2+1\) is irreducible over the reals, then \((x^2+1)^2\) contains that irreducible quadratic with multiplicity \(2\). If a squared quadratic itself factors, count the multiplicities of the factors after factoring it over the reals: \((x^2-2x+1)^2=(x-1)^4\), not an irreducible quadratic factor with multiplicity \(2\).
The denominator can be described in terms of linear and irreducible quadratic factors, each raised to its multiplicity. This is the real-factor viewpoint used in partial fractions. If numerator and denominator share a factor, cancel it before describing the factors of the simplified rational function; cancellation reduces the denominator multiplicity of that factor.
Worked Classifications
Worked Example: Classify a Denominator Already Written in Factors
Consider the proper rational function
The numerator has degree \(1\). The denominator has degree \(3+2(2)=7\), so \(1<7\) and the function is proper.
The factor \(x-2\) is linear and is raised to the third power. It is a linear factor with multiplicity \(3\).
For the quadratic \(3x^2-6x+12\), the coefficients are \(a=3\), \(b=-6\), and \(c=12\). Its discriminant is
Since \(-108<0\), this quadratic is irreducible over the reals. Completing the square gives
The expression \((x-1)^2+3\) is positive for every real \(x\), which confirms that this quadratic has no real root. It occurs to the second power, so the irreducible quadratic factor has multiplicity \(2\). The full denominator has degree \(3+2(2)=7\), as found above, confirming the degree count. The numerator’s only root is \(-\tfrac12\), which is neither \(2\) nor a root of the quadratic; no denominator factor cancels.
Worked Example: Factor Quadratics Before Counting Multiplicities
Classify the denominator of
The numerator has degree \(1\). The denominator has degree \(2(2)+2=6\), so \(1<6\) and the function is proper.
For \(x^2-5x+6\), the discriminant is
The roots are
so they are \(3\) and \(2\). Therefore \(x^2-5x+6=(x-3)(x-2)\). Since the original quadratic is squared, its factors each occur twice: \((x^2-5x+6)^2=(x-3)^2(x-2)^2\).
For \(2x^2+4x+10\),
This quadratic is irreducible over the reals. Completing the square confirms it:
Thus the denominator has the linear factors \(x-2\) and \(x-3\), each with multiplicity \(2\), and the irreducible quadratic \(x^2+2x+5\) with multiplicity \(1\). The constant factor \(2\) is not counted as a factor multiplicity. Reconstructing the denominator gives \(2(x-3)^2(x-2)^2((x+1)^2+4)\), which equals \((x^2-5x+6)^2(2x^2+4x+10)\). The degrees add to \(2+2+2=6\), as required. The numerator root is \(-4\); it is not \(2\) or \(3\), and the irreducible quadratic has no real root, so there is no cancellation.
Worked Example: A Repeated Root Hidden in a Quadratic
Classify the denominator of
The numerator has degree \(1\), and the denominator has degree \(2(2)+2(3)=10\). Hence \(1<10\), so the rational function is proper.
For the first quadratic,
Its repeated root is \(x=-b/(2a)=4/2=2\), so \(x^2-4x+4=(x-2)^2\). Because this factor is itself squared in the denominator, its full contribution is \(((x-2)^2)^2=(x-2)^4\). Thus the linear factor \(x-2\) has multiplicity \(4\).
For the second quadratic,
It is irreducible over the reals. Indeed, \(x^2+4x+8=(x+2)^2+4\), which is positive for every real \(x\). This irreducible quadratic has multiplicity \(3\). The denominator can therefore be written as \((x-2)^4(x^2+4x+8)^3\), with degree \(4+2(3)=10\), matching the original degree. The numerator’s root is \(-1\), which is not \(2\), and the irreducible quadratic has no real root, so no factor cancels.
A Common Discriminant Mistake
A frequent error is to use \(b^2+4ac\) instead of \(b^2-4ac\). For example, consider \(x^2+2x+5\). The incorrect calculation \(2^2+4(1)(5)=24\) might lead someone to claim that the quadratic has two real roots. But the discriminant is defined with a minus sign:
Therefore the quadratic has no real roots. Completing the square gives \(x^2+2x+5=(x+1)^2+4\), which is positive for every real \(x\). The factor is irreducible over the reals, not a product of real linear factors.
Another multiplicity error is to stop at the visible exponent. If a quadratic with a repeated root is already squared, factor the quadratic first and then apply the exponent: \((x^2-4x+4)^2=((x-2)^2)^2=(x-2)^4\). The linear factor has multiplicity \(4\), not \(2\).
Why Accurate Classification Helps
Partial fraction methods depend on the types and multiplicities of denominator factors. Misclassifying a quadratic changes the factor list; miscounting a power changes which denominator factors must be represented. A reliable process is therefore to factor over the reals, use the discriminant for each quadratic, count powers after factoring, and check that the degrees add back to the original denominator degree.
Compare the numerator and denominator degrees.
Compute \(b^2-4ac\), and factor it further when the discriminant is nonnegative.
Apply any outside power to the factors obtained after factoring the quadratic.
Reconstruct the denominator, compare degrees, and check for common factors with the numerator.
Check Your Understanding
For each question, classify the real factors and give their multiplicities where relevant.
- Is \(x^2-7x+12\) reducible over the reals or irreducible? Find its linear factors if it is reducible.
- Classify \(x^2+6x+13\) using its discriminant.
- Rewrite \((x^2-6x+9)^3\) using real linear factors, and state the multiplicity of its linear factor.
- For \(\dfrac{x+2}{(x^2+1)^2(x-3)^3}\), identify each denominator factor’s type and multiplicity. Is the function proper?
- What is the mistake in claiming that \(x^2+2x+5\) has two real roots because \(2^2+4(1)(5)>0\)?
